An inductor of inductance $L = 400 m H$ and resistors of resistances $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ are connected to a battery of emf $12 V$ as shown in figure. The internal resistance of the battery is negligible. The switch $S$ is closed at $t = 0$ . The potential drop across $L$ as a function of time is

\frac{12}{t} e^{- 3 t} V

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6 (1 - e^{- t / 0.2}) V

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12 e^{- 5 t} V

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6 e^{- 5 t} V

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Solution

The correct option is C $12 e^{- 5 t} V$
$I_{0}$ in LR circuit is
$I_{0} = \frac{12}{2} = 6$
$I = I_{0} (1 - e^{- 5 t})$
$τ = \frac{L}{R} = \frac{0.4}{2}$
$I = 6 (1 - e^{- 5 t})$
Voltage across inductor is $V_{L} = l \frac{d i}{d t} = 0.4 \times 6 \times 5 e^{- 5 t} = 12 e^{- 5 t}$

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Similar questions

Q. An inductor of inductance

L = 400 m H

and resistors of resistance

R_{1} = 2 Ω, R_{2} = 2 Ω

are connected to a battery of e.m.f

E = 12 V

as shown in the figure. The internal resistance of the battery is negligible. The switch

S

is closed at

t = 0

. The potential drop across

L

as a function of time is

An inductor of inductance L = 400 mH and resistors of resistance $R_{1} = 2 Ω$ and $R_{2} = 2 Ω$ areconnected to a battery of emf 12 V as shown in the figure. Theinternal resistance of the battery is negligible. The switch S is closed at t = 0. The potential drop across L as a function of time is